The curves $f(x)=\sin x$ and $g(x)=\cos x$ intersect periodically. Determine the area of the region bounded by these curves between $x=\dfrac{\pi}{4}$ and $x=\dfrac{5\pi}{4}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{2}}{2}$ (Choice B) B $2\sqrt{2}$ (Choice C) C $\dfrac{\sqrt{2}}{4}$ (Choice D) D $\sqrt{2}$
Visualizing the area We sketch the curves $f(x)=\sin x$ and $g(x)=\cos x$ first. ${1}$ ${2}$ ${3}$ ${4}$ ${1}$ ${\llap{-}1}$ $g$ $f$ $y$ $x$ From the graph, it appears that $f(x) \ge g(x)$ between $x=\dfrac{\pi}4$ and $x=\dfrac{5\pi}4$. From this we are looking to evaluate: $ \int_{{\pi}/{4}}^{{5\pi}/{4}}\left( f(x)-g(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=}\int_{\pi/4}^{5\pi/4}\left(\sin x-\cos x\right)\,dx \\\\ &=\left(-\cos x-\sin x\right)~\Bigg|_{\pi/4}^{5\pi/4} \\\\ &=-\left(\cos x+\sin x\right)~\Bigg|_{\pi/4}^{5\pi/4} \\\\ &=-\left(-\dfrac{\sqrt2}{2}-\dfrac{\sqrt2}{2}\right)+\left(\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\right) \\\\ &= 2\sqrt2 \end{aligned}$ Answer The area is $2\sqrt{2}$ square units.